What mass of \( NaCl (58.5 g / mol ) \) must be dissolved in \( 65 g \

1.	What mass of \( NaCl (58.5 g / mol ) \) must be dissolved in \( 65 g \) water to lower the freezing point by \( 7.5^{\circ} C \) ? \( Kr ^{-1.86} \) \( K Kg / mol \) and \( i \) for \( Na Cl =1.86 \).
Answer» As we know \(\Delta T_f = iK_fm\) i = Van't hoff factor ⇒ \(7.5 ^\circ K = 1.86 \times 1.86 \times \frac{\frac{w}{58.5}\times 1000}{65}\) ⇒ \(w = \frac{7.5 \times 65 \times 58.5}{1.86 \times 1.86 \times 1000}\) w = 0.24 g

What mass of \( NaCl (58.5 g / mol ) \) must be dissolved in \( 65 g \) water to lower the freezing point by \( 7.5^{\circ} C \) ? \( Kr ^{-1.86} \) \( K Kg / mol \) and \( i \) for \( Na Cl =1.86 \).

Answer»

As we know

\(\Delta T_f = iK_fm\)

i = Van't hoff factor

⇒ \(7.5 ^\circ K = 1.86 \times 1.86 \times \frac{\frac{w}{58.5}\times 1000}{65}\)

⇒ \(w = \frac{7.5 \times 65 \times 58.5}{1.86 \times 1.86 \times 1000}\)

w = 0.24 g

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What mass of \( NaCl (58.5 g / mol ) \) must be dissolved in \( 65 g \) water to lower the freezing point by \( 7.5^{\circ} C \) ? \( Kr ^{-1.86} \) \( K Kg / mol \) and \( i \) for \( Na Cl =1.86 \).

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