Home
Class 12
CHEMISTRY
By dissolving 13.6 g of a substance in 2...

By dissolving `13.6 g` of a substance in `20 g` of water, the freezing point decreased by `3.7^(@)C`. Calculate the molecular mass of the substance. (Molal depression constant for water `= 1.863K kg mol^(-1))`

Text Solution

AI Generated Solution

To calculate the molecular mass of the substance based on the given data, we can follow these steps: ### Step 1: Identify the given values - Mass of the solute (substance) \( W_2 = 13.6 \, \text{g} \) - Mass of the solvent (water) \( W_1 = 20 \, \text{g} \) - Depression in freezing point \( \Delta T_f = 3.7 \, ^\circ C \) - Molal depression constant for water \( K_f = 1.863 \, \text{K kg mol}^{-1} \) ...
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    CENGAGE CHEMISTRY ENGLISH|Exercise Solved Examples|40 Videos

  • SOLUTIONS

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises (Linked Comprehension)|58 Videos

  • SOLID STATE

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 1.2 (Objective)|9 Videos

  • SURFACE CHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Archives Subjective|2 Videos

Similar Questions

Explore conceptually related problems

5 g of a substance when dissolved in 50 g water lowers the freezing by 1.2^(@)C . Calculate molecular wt. of the substance if molal depression constant of water is 1.86 K kg mol^-1 .

A solution containing 12.5g of a non electrolyte substance in 175g of water gave a boiling point elevation of 0.70 K. Calculate the molar mass of the substance. (Elevation constant for water, K_(b) = 0.52 K kg "mol"^(-1) )

6 g of a substance is dissolved in 100 g of water depresses the freezing point by 0.93^(@)C . The molecular mass of the substance will be: ( K_(f) for water = 1.86^(@)C /molal)

A solution containg 12 g of a non-electrolyte substance in 52 g of water gave boiling point elevation of 0.40 K . Calculate the molar mass of the substance. (K_(b) for water = 0.52 K kg mol^(-1))

On dissolving 0.25 g of a non-volatile substance in 30 mL benzene (density 0.8 g mL^(-1)) , its freezing point decreases by 0.25^(@)C . Calculate the molecular mass of non-volatile substance (K_(f) = 5.1 K kg mol^(-1)) .

0.48 g of a substane was dissolved in 10.6g C_(6)H_(6) . The freezing point of benzene was lowered by 1.8^(@)C . Calculate m.w.t. of the substance. Molecular depression constant for benzene is 50 K"mol"^(-1) 100 g .

1.355 g of a substance dissolved in 55 g of CH_(3)COOH produced a depression in the freezing point of 0.618^(@)C . Calculate the molecular weight of the substance (K_(f)=3.85)

When 10.6 g of a non volatile substance is dissolved in 750 g of ether, its boiling point is raised 0.266^(@)C . The molecular weight of the substance is (Given : Molal boiling point constant for ether is 2.0^(@)Ckg//mol ) Report your answer by rounding it up to nearest whole number.

2.71 g of a substance dissolved in 55 g of CH2COOH produced a depression in the freezing point of 0.618 o C. Calculate the molecular weight of the substance. (Kf=3.85)

A solution containing 12.5 g of non-electrolyte substance in 185 g of water shows boiling point elevation of 0.80 K. Calculate the molar mass of the substance. ( K_b=0.52 K kg mol^(-1) )