How much amount of NaCl should be added to 500 g of water ( ρ =1.00g/mL ) to decrease the freezing point of water to -0.3 ° C? (The freezing point depression constant for water =2 Kkgmol- 1 )

Question

How much amount of NaCl should be added to 500 g of water ( ρ =1.00g/mL ) to decrease the freezing point of water to -0.3 ° C? (The freezing point depression constant for water =2 Kkgmol- 1 ) (A) 2.19 g (B) 1.88 g (C) 1.96 g (D) 1.085 g

Tardigrade Admin · Accepted Answer

Δ Tf=0.3 Δ Tf=iKfm i=2 for NaCl 0.3=2× 2× m m=(w/58.5)× (1000/500) 0.3=2× 2× (w/58.5)× (1000/500) w=(0.3 × 58.5 × 500/4 × 1000)=2.19g

Q. How much amount of NaCl should be added to 500 g of water ( $ρ = 1.00 g / m L$ ) to decrease the freezing point of water to -0.3 $^{\circ}$ C?
(The freezing point depression constant for water $= 2 K k g m o l^{- 1}$ )

2.19 g

1.88 g

1.96 g

1.085 g

$Δ T_{f} = 0.3$
$Δ T_{f} = i K_{f} m$
$i = 2$ for NaCl
$0.3 = 2 \times 2 \times m$
$m = \frac{w}{58.5} \times \frac{1000}{500}$
$0.3 = 2 \times 2 \times \frac{w}{58.5} \times \frac{1000}{500}$
$w = \frac{0.3 \times 58.5 \times 500}{4 \times 1000} = 2.19 g$

Q. How much amount of NaCl should be added to 500 g of water ( ρ=1.00g/mL ) to decrease the freezing point of water to -0.3 ∘ C? (The freezing point depression constant for water =2Kkgmol−1 )