The amount of urea to be dissolved in $500$ cc of water $(K_{f} = 1.86)$ to produce a depression of ${0.186}^{o} C$ in the freezing point is:

9

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6

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3

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5

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Solution

The correct option is C $3$ g
We know,
$△ T = K_{f} \times \frac{w_{B} \times 1000}{M_{B} \times w_{A}}$

$0.186 = 1.86 \times \frac{w_{B} \times 1000}{60 \times 500}$

$w_{B} = 3 g$

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Similar questions

The amount of Urea to be dissolved is 500cc of water ( $K_{f}$ =1.86) to produce a depression of ${0.186}^{\circ} C$ in the freezing point is

500 c c

(K_{f} = {1.86}^{\circ} C k g m o l^{- 1})

{0.186}^{\circ} C

60 g / m o l

The amount of urea to be dissolved in 500 ml of water (K =18.6 K mole^-1 in 100g solvent) to produce a depression of 0.186°C in freezing point is

500 m l

0.186 K

K_{f} = 1.86 K k g m o l^{- 1} ρ_{H_{2} O} = 1 g m L^{- 1}

When 5.85 g of NaCl and 6 g of urea are dissolved in 1000 g of water, the freezing point gets depressed by $K_{f}$ for water is 1.86

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