150 g of acetic acid was contaminated with 10.2 g ascorbic acid () to lower down its freezing point by . The value of x is (Nearest integer).

[Given: ; Molar mass of ascorbic acid = 176 g mol⁻¹]

Question

150 g of acetic acid was contaminated with 10.2 g ascorbic acid () to lower down its freezing point by . The value of x is  (Nearest integer).

[Given: ; Molar mass of ascorbic acid = 176 g mol⁻¹]

Canvas Classes · Accepted Answer

Strategy:
 Find the molality of the solution, then calculate the freezing point depression using . Finally, match the result to the form  to find .

Step 1: Calculate moles of ascorbic acid ()
Molar mass = 176 g/mol. Mass = 10.2 g.

Step 2: Calculate the molality
Mass of acetic acid (solvent) = 150 g = 0.15 kg.

Step 3: Calculate 
.

Step 4: Find } x
.
Rounding to the nearest integer, .

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