Question

what weight of glycerol should be added to
600 g of water in order to lower its
freezing point by 10 c ? (kf= 1.86 c/m)​

Answer 1

Answer:

ΔTf = 100C ΔTf = Kf × molality 10 = 1.86 × x × 100092.09 × 600 x = 10 × 600 × 92.091.86 × 1000 = 297.06g Wt. of glycerol taken = 297.06g.

Answer 2

Answer:

The weight of glycerol should be added in 600g of water to lower its freezing point by 10°C, is equal to 296.8g.

Explanation:

We have given, the lowering in freezing point, $\triangle T_{f}= 10^{o}C$

The value of molal freezing depression constant, $K_{f}= 1.86^oCm^{-1}$

We know that, $\triangle T_{f}= K_{f}.m$                                      ........................(1)

where m is molality.

Given , the weight of water (solvent), W₂ = 600g = 0.6Kg

Molality, $m=\frac{W_{1}}{M*W_{2}}$

where W₁ is the weight of glycerol,

M is the molecular mass of glycerol = 92gmol⁻¹

So, $m= \frac{W_{1}}{92*(0.6)}$

m = 0.0181×W₁

Put all the value  in equation (1);

$10=1.86*(0.01811*W_{1})$

W₁ = 296.8g

Therefore, the weight of glycerol equals to 296.8g, should be added in water.