How much amount of NaCl should be added to 600 g of water ( ρ =1.00 g / m L ) to decrease the freezing point of water to -0.2oC ? (The freezing point depression constant for water = text2 K kg mo textl- 1 )

Question

How much amount of NaCl should be added to 600 g of water ( ρ =1.00 g / m L ) to decrease the freezing point of water to -0.2oC ? (The freezing point depression constant for water = text2 K kg mo textl- 1 ) (A) 2.14 g (B) 0.88 g (C) 1.96 g (D) 1.76 g

Tardigrade Admin · Accepted Answer

Δ Tf=0.2 Δ Tf=iKfm i=2 for NaCl 0.2=2× 2× m m=(w/58.5)× (1000/600) 0.2=2× 2× (w/58.5)× (1000/600) w=(0.2 × 58.5 × 600/4 × 600)=(1.2 × 58.5/40)=1.76g

Q. How much amount of NaCl should be added to 600 g of water ( $ρ = 1.00 g / m L$ ) to decrease the freezing point of water to $- 0. 2^{o} C$ ?

(The freezing point depression constant for water $= 2 K kg mo l^{- 1}$ )

2.14 g

0.88 g

1.96 g

1.76 g

$Δ T_{f} = 0.2$

$Δ T_{f} = i K_{f} m$

$i = 2$ for NaCl

$0.2 = 2 \times 2 \times m$

$m = \frac{w}{58.5} \times \frac{1000}{600}$

$0.2 = 2 \times 2 \times \frac{w}{58.5} \times \frac{1000}{600}$

$w = \frac{0.2 \times 58.5 \times 600}{4 \times 600} = \frac{1.2 \times 58.5}{40} = 1.76 g$

Q. How much amount of NaCl should be added to 600 g of water ( ρ=1.00g/mL ) to decrease the freezing point of water to −0.2oC ? (The freezing point depression constant for water =2 K kg mol−1 )