Home
Class 11
CHEMISTRY
500 gms of water contains 6 times 10^(-3...

500 gms of water contains `6 times 10^(-3)` gms of dissolved `MgSO_(4)` in it. Calculate the hardness of water in ppm of `CaCO_(3)`?

A

20 ppm

B

30 ppm

C

40 ppm

D

10 ppm

Text Solution

AI Generated Solution

The correct Answer is:
To calculate the hardness of water in ppm of CaCO₃ based on the given amount of dissolved MgSO₄, follow these steps: ### Step 1: Identify the given data - Mass of water = 500 g - Mass of dissolved MgSO₄ = 6 x 10^(-3) g ### Step 2: Calculate the molecular weight of MgSO₄ - The molecular weight of MgSO₄ can be calculated as: - Magnesium (Mg) = 24 g/mol - Sulfur (S) = 32 g/mol - Oxygen (O) = 16 g/mol (there are 4 oxygen atoms in MgSO₄) Thus, the molecular weight of MgSO₄ = 24 + 32 + (16 x 4) = 120 g/mol. ### Step 3: Calculate the equivalent mass of CaCO₃ for the given mass of MgSO₄ - The molecular weight of CaCO₃ is: - Calcium (Ca) = 40 g/mol - Carbon (C) = 12 g/mol - Oxygen (O) = 16 g/mol (there are 3 oxygen atoms in CaCO₃) Thus, the molecular weight of CaCO₃ = 40 + 12 + (16 x 3) = 100 g/mol. ### Step 4: Set up the proportion to find the equivalent mass of CaCO₃ - Using the ratio of the molecular weights, we can set up the equation: \[ \frac{6 \times 10^{-3} \text{ g MgSO₄}}{120 \text{ g MgSO₄}} = \frac{x \text{ g CaCO₃}}{100 \text{ g CaCO₃}} \] ### Step 5: Solve for x (mass of CaCO₃) - Cross-multiplying gives: \[ x = \frac{6 \times 10^{-3} \times 100}{120} \] - Simplifying this: \[ x = \frac{6 \times 10^{-1}}{120} = 5 \times 10^{-3} \text{ g CaCO₃} \] ### Step 6: Calculate the hardness in ppm - The formula for hardness in ppm is given by: \[ \text{Hardness (ppm)} = \left( \frac{\text{mass of CaCO₃}}{\text{mass of water sample}} \right) \times 10^6 \] - Substituting the values: \[ \text{Hardness (ppm)} = \left( \frac{5 \times 10^{-3} \text{ g}}{500 \text{ g}} \right) \times 10^6 \] ### Step 7: Calculate the final result - Performing the calculation: \[ \text{Hardness (ppm)} = \left( \frac{5 \times 10^{-3}}{500} \right) \times 10^6 = \left( 1 \times 10^{-5} \right) \times 10^6 = 10 \text{ ppm} \] ### Final Answer The hardness of water in ppm of CaCO₃ is **10 ppm**. ---

To calculate the hardness of water in ppm of CaCO₃ based on the given amount of dissolved MgSO₄, follow these steps: ### Step 1: Identify the given data - Mass of water = 500 g - Mass of dissolved MgSO₄ = 6 x 10^(-3) g ### Step 2: Calculate the molecular weight of MgSO₄ - The molecular weight of MgSO₄ can be calculated as: ...
Promotional Banner

Topper's Solved these Questions

  • HYDROGEN & ITS COMPOUNDS

    NARAYNA|Exercise CHECK YOUR GRASP|7 Videos
  • HYDROGEN & ITS COMPOUNDS

    NARAYNA|Exercise EVALUATE YOURSELF - 1|5 Videos
  • HYDROGEN & ITS COMPOUCDS

    NARAYNA|Exercise Interger type questions|10 Videos
  • IONIC EQUILIBRIUM

    NARAYNA|Exercise Level-VI (H.W)|35 Videos

Similar Questions

Explore conceptually related problems

Solution of 100 ml water contains 0.73 g of Mg(HCO_(3))_(2) and 0.81g of Ca(HCO_(3))_(2) calculate the hardness ini terms of ppm of CaCO_(3)

A sample of hard water contains 1 mg CaCl_2 and 1 mg MgCl_2 per litre. Calculate the hardness of water in terms of CaCO_3 present in per 10^(6) parts of water. (a). 2.5 ppm (b). 1.95 ppm (c). 2.15 ppm (d). 195 ppm

100g of a water samples is found to contain 12 mg of MgSO_(4) calculate the hardness of water sample.

On litre of sea water weighs 1050 grams and contains 6 xx 10^(-3)g of dissolved oxygen gas. Calculate the concentration of the dissolved oxygen in ppm.

One litre of a sample of hard water contains 1 mg of CaCl_(2) and 1 mg of MgCl_(2) . Find the total hardness of water in terms of parts of CaCO_(3) per 10^(6) parts of water by mass.

One litre of a sample of hard water contain 4.44mg CaCl_(2) and 1.9mg "of" MgCl_(2) . What is the total hardness in terms of ppm of CaCO_(3)?

A water is said to be soft water if it produces sufficient foam with the soap and water that does not produce foam with soap is known as hard water. Hardness has been classified into two types (i)Temporary hardness (ii) Permanent hardness. Temporary hardness is due to presence of calcium and magnesium bicarbonate. It is simply removed by boiling as Ca(HCO_(3))_(2)overset(Delta)rarr CaCO_(3)darr+CO_(2)uarr+H_(2)O Mg(HCO_(3))_(2)overset(Delta)rarr MgCO_(3)darr+CO_(2)uarr+H_(2)O temporary hardness can also be removed by addition of slaked lime, Ca(OH)_(2) Ca(HCO_(3))_(2)+Ca(OH)_(2) to 2CaCO_(3)darr+2H_(2)O permanent hardsness is due to presencce of sulphates and chlorides of Ca,Mg,etc. It is removed by washing soda as CaCl_(2)+Na_(2)CO_(3) to CaCO_(3)darr+2NaCl CaSO(4)+Na_(2)CO_(3)to CaCO_(3)darr+Na_(2)SO_(4) Permanent hardness also removed by ion exchange resin process as 2RH+Ca^(2+)toR_(2)Ca+2H^(+) 2ROH+SO_(4)^(2-) to R_(2)SO_(4)+2OH^(-) The degree of hardness of water is measured in terms of PPm of CaCO_(3) 100 PPm means 100 g of CaCO_(3) is present in 10^(6) g of H_(2)O . If any other water sample which contain 120 PPm of MgSO_(4) , hardness in terms of CaCO_(3) is equal to =100 PPm. One litre of a sample of hard water (d=1 g/mL) cotains 136 mg of CaSO_(4) and 190 mg of MgCl_(2) . What is the total hardness of water in terms of CaCO_(3) ?

100 mL sample of hard water is titrated with 500 " mL of " 0.001 M EDTA solution at pH=10 , using eriochrome black-T indicator to reach equivalence point. An equal another amount of hard water sample is boiled for 30 min. After filtration and cooling, the same sample is titrated with 200 " mL of " 0.011 M EDTA solution at pH=10 using Mg-EDTA complex solution and erichrome black-T indicator to reach equivalence point. (i). Calculate the total hardness of water sample (temporary + permanent) in ppm of CaCO_3 . (ii). Calculate the permanent hardness of water sample in ppm of CaCO_3 . (iii). Calculate the temporary hardness of water sample in ppm of CaCO_3

10 L of hard water requires 0.28 g of line (CaO) for removing hardness. Calculate the temporary hardness in ppm of CaCO_(3) .

NARAYNA-HYDROGEN & ITS COMPOUNDS-LEVEL-4
  1. 500 gms of water contains 6 times 10^(-3) gms of dissolved MgSO(4) in ...

    Text Solution

    |

  2. Hydrogen resembles halogens in many respects for which several fac...

    Text Solution

    |

  3. Why does H^(+) ion always get associated with atoms or molecules ...

    Text Solution

    |

  4. When zeolite which is hydrated sodium aluminium silicate is treated wi...

    Text Solution

    |

  5. Which of the following hydrides is electron-precise hydride ?

    Text Solution

    |

  6. Elements of which of the following group(s) of periodic table do ...

    Text Solution

    |

  7. Only one element of …….from hydride.

    Text Solution

    |

  8. Metal hydrides are ionic, covalent or molecular in nature. Among LiH, ...

    Text Solution

    |

  9. Which of the following statements is incorrect?

    Text Solution

    |

  10. The oxide that give H(2)O(2) on treatment with dilute H(2)SO(4) is

    Text Solution

    |

  11. Hydrogen peroxide is obtained by the electrolysis of ………

    Text Solution

    |

  12. Which of the following equations depict theoxidising nature of H(2)O(2...

    Text Solution

    |

  13. Which of the following equation depicts reducing nature of H(2)O(2)...

    Text Solution

    |

  14. Cosider the reactions (i) H(2)O(2) + 2HI to I(2) + 2H(2)O (ii...

    Text Solution

    |

  15. Which of the following reaction produce dihydrogen from synthesis gas?

    Text Solution

    |

  16. When sodium peroxide is treated with dilute H(2)SO(4), we get

    Text Solution

    |

  17. Which of the following compounds is used to remove water hardens

    Text Solution

    |

  18. When zeolite which is hydrated sodium aluminium silicate is treated wi...

    Text Solution

    |

  19. In which of the following reactions does hydrogen acts as an oxidisng ...

    Text Solution

    |

  20. CO+H(2) underset("catalyst")overset(300^(@)//300atm)(rarr)CH(3)OH the ...

    Text Solution

    |

  21. Which one of the following statements regarding H(2)O(2) is false?

    Text Solution

    |