Answer is: the approximate freezing point of a 0.10 m NaCl solution is -2x°C.
V<span>an't Hoff factor (i) for NaCl solution is approximately 2.
</span>Van't Hoff factor (i) for glucose solution is 1.<span>
Change in freezing point from pure solvent to solution: ΔT = i · Kf · m.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
m - molality, moles of solute per kilogram of solvent.
</span>Kf and molality for this two solutions are the same, but Van't Hoff factor for sodium chloride is twice bigger, so freezing point is twice bigger.

8 0

3 years ago

The empirical formula of a group of compounds is CHCl. Lindane, a powerful insecticide, is a member of this group. The molar mas

Sergeeva-Olga [200]

Answer:

Explanation:

since, n = molar mass / empirical formula mass

Empirical formula mass = Total mass of atoms present in empirical formula

CHCl = 12+1+35.5

= 48.5

Given, Molar mass = 290.8 g.

So, n = 290.8/48.5

= 5.995 , that is approx 6.

So, Molecular formula = n × Empirical formula

= 6 × CHCl

= $C_{6}H_{6}Cl_{6}$

So, Number of C = 6

3 0

3 years ago