Name: Osmosis and Osmotic Pressure
Uploaded: 2019-07-19
Duration: 7 min 24 s
Description: What does "RO" in your water filter stand for? This video introduces the concepts of osmosis, reverse osmosis and osmotic pressure. Also, learn how to calcul...

Question

A

5 %

solution of cane sugar in water has a freezing point of

271 K

. Calculate the freezing point of

5 %

(by mass) solution of glucose in water. The freezing point of pure water is

273.15 K

. (Molar mass of cane sugar

= 342 g {mol}^{- 1}

, molar mass of glucose

= 180 g {mol}^{- 1}

)

Accepted Answer

Given,

The freezing point of the solution of cane sugar $271 K$

The freezing point of pure water $= 273.15 K$

The depression in the freezing point of the glucose solution $∆ T_{f (sucrose)} = 273.15 - 271 = 2.15 K$

A $5 %$ solution of sucrose by mass and $5 %$ solution of by mass means that

The weight of the sucrose $W_{(sucrose)} = 5 g$

The molar mass of sucrose $M_{(sucrose)} = 348 g {mol}^{- 1}$

The weight of the glucose $W_{(glucose)} = 5 g$

The molar mass of the glucose $M_{(glucose)} = 180 g {mol}^{- 1}$

The weigh of the water for both solutions $W_{(water)} = 95 g$

So, The depression in the freezing point for the sucrose solution

$\Rightarrow ∆ T_{f (sucrose)} = K_{f} \times \frac{W_{(sucrose)} \times 1000}{M_{(sucrose)} \times W_{(water)}} . . . . . . . . . . (1)$

The depression in freezing point for the glucose solution

$\Rightarrow ∆ T_{f} = K_{f} \times \frac{W_{(glucose)} \times 1000}{M_{(glucose)} \times W_{(water)}} . . . . . . . . . . (2)$

By dividing both equations

$\Rightarrow \frac{∆ T_{f (sucrose)}}{∆ T_{f (glucose)}} = \frac{\frac{W_{(sucrose)} \times 1000}{M_{(sucrose)} \times W_{(water)}}}{\frac{W_{(glucose)} \times 1000}{M_{(glucose)} \times W_{(water)}}}$

$\Rightarrow \frac{2.15}{∆ T_{f (glucose)}} = \frac{\frac{5 \times 1000}{342 \times 95}}{\frac{5 \times 1000}{180 \times 95}} \Rightarrow \frac{2.15}{∆ T_{f (glucose)}} = \frac{180}{342} \Rightarrow ∆ T_{f (glucose)} = \frac{342 \times 2.15}{180} = 4.085$

So, the freezing point of the glucose solution $= T_{f (water)} - ∆ T_{f (glucose)} = 273.15 - 4.085 = 269.065 K$ .

A $5 %$ solution of cane sugar in water has a freezing point of $271 K$ . Calculate the freezing point of $5 %$ (by mass) solution of glucose in water. The freezing point of pure water is $273.15 K$ . (Molar mass of cane sugar $= 342 g {mol}^{- 1}$ , molar mass of glucose $= 180 g {mol}^{- 1}$ )

Important Questions on Solutions

Learn and Prepare for any exam you want !

A 5% solution of cane sugar in water has a freezing point of 271 K. Calculate the freezing point of 5% (by mass) solution of glucose in water. The freezing point of pure water is 273.15 K. (Molar mass of cane sugar =342 g mol-1, molar mass of glucose =180 g mol-1)

Important Questions on Solutions

Learn and Prepare for any exam you want !

A $5 %$ solution of cane sugar in water has a freezing point of $271 K$ . Calculate the freezing point of $5 %$ (by mass) solution of glucose in water. The freezing point of pure water is $273.15 K$ . (Molar mass of cane sugar $= 342 g {mol}^{- 1}$ , molar mass of glucose $= 180 g {mol}^{- 1}$ )