Question

A mixture of KCl and KNO3

Answer 1

Answer is: The percent of KCl in the mixture is closest to a) 40%.

If we use 100 grams of mixture:
ω(K) = 44.20% ÷ 100% = 0.442.
m(K) = 0.442 · 100 g = 44.2 g.
n(K) = 44.2 g ÷ 39.1 g/mol.
n(K) = 1.13 mol.
n(KCl) + n(KNO₃) = n(K)
m(KCl) = x.
m(KNO₃) = y.
Two equtions:
1) x + y = 100 g.
2) m(KCl)/M(KCl) + m(KNO₃)/M(KNO₃) = 1.13 mol.
x/74.55 g/mol + y/101.1 g/mol = 1.13 mol.
From first equation find x = 100 - y and put in second equation:
(100 - y)/74.55 + y/101.1 = 1.13 /×101.1.
135.61 - 1.356y + y = 114.24.
0.356y = 22.37 g.
y = 62.83 g.
x = m(KCl) = 100 g - 62.83 g = 37.17 g.

Answer 2

Answer:

The correct answer is option (a).

Explanation:

Suppose we have total mass of the mixture be 100 g

let the mass of KCl be x

let the mass of $KNO_3$ be y

x + y = 100...(1)

Moles of KCl = $\frac{x}{75.5 g/mol}$

Moles of K in KCl will be = $\frac{x}{75.5 g/mol}$ (1 mol is present in 1 mole of molecule)

Moles of $KNO_3=\frac{y}{102 g/mol}$

Moles of K in $KNO_3$ will be = $\frac{y}{102 g/mol}$ (1 mol is present in 1 mole of molecule)

Percentage of potassium in the mixture = 44.20 %

In 100 g, the mass of potassium = 44.20 g

Moles of potassium =$\frac{44.20 g}{40 g/mol}=1.105 mol$

So,

$\frac{x}{75.5 g/mol}+\frac{y}{102 g/mol}=1.105 mol$..(2)

Solving equation (1) and (2):

we get ,y = 63.78 g, x=36.22 g

Percentage of KCl:

$\frac{36.22 g}{100g}\times 100=36.22 g\%$

The percent of KCl in the  mixture is closest to 40%.Hence ,option (a) is correct.