Home
Class 12
CHEMISTRY
5 g of a substance when dissolved in 50 ...

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`.

Text Solution

AI Generated Solution

To solve the problem step by step, we will use the formula for depression in freezing point, which is derived from Raoult's law. ### Step 1: Understand the given data - Mass of the solute (substance) = 5 g - Mass of the solvent (water) = 50 g = 0.050 kg (since we need it in kg for molality) - Depression in freezing point (ΔTf) = 1.2 °C - Molal depression constant of water (Kf) = 1.86 K kg mol⁻¹ ...
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    AAKASH INSTITUTE ENGLISH|Exercise TRY YOURSELF|14 Videos

  • SOLUTIONS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGMENT (SECTION-A)|50 Videos

  • REDOX REACTIONS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT SECTION - D|10 Videos

  • SOME BASIC CONCEPT OF CHEMISTRY

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT( SECTION - D) Assertion-Reason Type Questions|15 Videos

Similar Questions

Explore conceptually related problems

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))

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)

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 .

What mass of NaCl ("molar mass" =58.5g mol^(-1)) be dissolved in 65g of water to lower the freezing point by 7.5^(@)C ? The freezing point depression constant K_(f) , for water is 1.86 K kg mol^(-1) . Assume van't Hoff factor for NaCl is 1.87 .

What mass of NaCI ("molar mass" =58.5g mol^(-1)) be dissolved in 65g of water to tower the freezing point by 7.5^(@)C ? The freezing point depression constant, K_(f) , for water is 1.86 K kg mol^(-1) . Assume van't Hoff factor for NaCI is 1.87 .

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)

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)

1.5g of monobasic acid when dissolved in 150g of water lowers the freezing point by 0.165^(@)C.0.5g of the same acid when titrated, after dissolution in water, requires 37.5 mL of N//10 alkali. Calculate the degree of dissociation of the acid (K_(f) for water = 1.86^(@)C mol^(-1)) .

Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol^(-1). The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is 46 g mol^(-1) Among the following, the option representing change in the freezing point is

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) )