At what temperature would a 1.65 m NaCl solution freeze, given that the van't Hoff factor for NaCl is 1.9? Kf for water is 1.86 ∘C/m .

-5.83 °C

Explanation:

Step 1: Given data

Step 2: Calculate the freezing point depression

The freezing point depression is a colligative property that can be calculated using the following expression.

ΔTf = i × m × Kf

ΔTf = 1.9 × 1.65 m × 1.86 °C/m

ΔTf = 5.83 °C

Step 3: Calculate the freezing point of the solution

The freezing point of pure water is 0°C. The freezing point of the solution is:

0 °C - 5.83°C = -5.83 °C

i'm assuming that the correct answer is indeed the first one/ "a"..

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2