Home
Class 11
CHEMISTRY
A sample of hard water is found to conta...

A sample of hard water is found to contain 50mg of `CaSO_4` in 12kg of sample.Calculate the hardness of the sample in terms of ppm of `CaSO_4`.

Promotional Banner

Topper's Solved these Questions

  • STRUCTURE OF ATOM

    ICSE|Exercise Objective (MCQ) TYPE Questions|67 Videos

  • STRUCTURE OF ATOM

    ICSE|Exercise True Or False Type Questions|19 Videos

  • STRUCTURE OF ATOM

    ICSE|Exercise Short Answer Type Questions|30 Videos

  • STATES OF MATTER : GASES AND LIQUIDS

    ICSE|Exercise NCERT TEXT-BOOK EXERCISES (With Hints and Solutions)|23 Videos

  • THE s - BLOCK ELEMENTS

    ICSE|Exercise NCERT TEXT-BOOK. EXERCISES (WITH HINTS AND SOLUTIONS)|55 Videos

Similar Questions

Explore conceptually related problems

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

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 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) ?

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 in terms of ppm of CaCO_(3)

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)?

One litre of sample of hard water contain 4.44mg CaCI_2 and 1.9mg of MgCI_2 . What is the total hardness in terms of ppm of CaCO_3 ?

A sample of hard water is found to contain 40 mg of Ca^(2+) ions per litre. The amount of washing shoda (Na_(2)CO_(3)) required to soften 5*0 L of the sample would be

One litre of sample of hard water contains 0.8 mg of CaCl_(2) and "0.8 mg of MgCl"_(2) . Find the total hardness in terms of parts of CaCO_(3) , per 10^(6) parts of water by mass.

A sample of sucrose is found to contain 72.28 xx 10^21 atoms of carbon. Find the mass of the sample in grams.

A 10 mL sample of human urine was found to have 5 mg of urea on analysis. Calculate the molarity of the given sample w.r.t. urea. (molecular mass of urea = 60)