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2.82 g of glucose (molar mass = 180) is ...

2.82 g of glucose (molar mass = 180) is dissolved in 30 g of water. Calculate (a) the molality.

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To calculate the molality of the solution, we will follow these steps: ### Step 1: Calculate the number of moles of glucose To find the number of moles of glucose, we use the formula: \[ \text{Number of moles} = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)}} \] Given: ...
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2.82g of glucose (molar mass =180 ) is dissolved in 30g of water. Calculate the (i) Molality of the solution (ii) mole fractions of (a) glucose (b) water.

2.82 g of glucose (molar mass = 180) is dissolved in 30 g of water. Calculate (b) mole fraction of glucose and water.

Knowledge Check

  • 1.80 g of glucose (molar mas =180) is dissolved in 36.0 g of water in a beaker. The total number of oxygen atoms in the solution is

    A
    `12.405xx10^(23)`
    B
    `12.405xx10^(22)`
    C
    `6.022xx10^(23)`
    D
    `6.022xx10^(22)`

  • If 3g of glucose (molecular mass 180) is dissolved in 60g of water at 15^(@)C , then the osmotic pressure of this solution will be :

    A
    0.34 atm
    B
    0.65 atm
    C
    6.57 atm
    D
    5.57 atm

  • If 23 g of glucose ( molecular mass 180)is dissolved in 60ml of water at 15^(@)C , then the osmotic pressure of this solution will be

    A
    0.34 atm
    B
    0.75 atm
    C
    50.35 atm
    D
    5.57 atm.

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