Question

A solid consists of a mixture of NaNO3 and Mg(NO3)2. When 6.50 g of the solid is dissolved in 50.0 g of water, the freezing point of the solution is lowered by 5.06°C. What is the composition by mass of the solid?

____ g NaNO₃ ?
____ g Mg(NO₃)₂ ?

Answer 1

We know, AT =kem (Where me molality KF - Freezing point constant T AT = temp. difference Yession constant (ke) for Water as a solvent is '7.86°C/m 5.06 = 1.86 mm molality = 2.727 That means 2.72 moles of ions in 1000 g of Water Moles of ion in 50g of Water - = 2.72 x 5o = 0.136 moles y Toon* 50 = 0.136 mole ions Now, we know 1 NaNO molecule releases 2 ions 1 Mg(NO3) molecule release Zious & Let weight of Nano in given mixture be a gram So weight of My (10) will be (6.5-x.) gram Number of moles of ions produced by c gram g Na NO x2 = 0.0 235 x * Number of moles of ions produced by (6.5-23 gram = 148.3 79.5-34 148.3 - 0.131 -- 0.02. se!

As Calculated earlier , Total moles of ions in 50g of Water = 0.136. This must be equal to sum of moles of ions produced by NaNO, & Mg(NO3)2 → Hence, 0.136 = 0.02352 + (0.13/-0.02%) 0.136 -0.13) =(0.0235 -0.02) * x = 0.005 0.0035 = 1.428 g * mass of NaNO = 1.428 g. * mass of Mo (NO₂) = (6.5-1.428) 32 = 5.072 q. 3 in term of percentage 1.428 x 100 - 21.9% of Na NO 6.5 5.072 x 100 = 78.1 % og Mg(NOx), 6.5

Answer:

NaNO₃ : 1.428 g

Mg(NO_{​​​​​3}​)₂ : 5.072 g