Question

% (w/v) of aqueous KOH solution is 1.4. If the molarity of
solution is a M then value of 100a is (Molar mass of KOH =
56​

Answer 1

Answer:

100 g solution contains 30 g KOH

Therefore ,

d

100

mL solution contains

56

30

mol KOH , d=density in g/m

2

Molarity =6.90=

56

30

×

100

d

×1000

d=1.288 g/mL.

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Answer 2

Complete Question:

% (w/v) of aqueous KOH solution is 1.4. What is the molarity of the solution? (Molar mass of KOH = 56)​

Answer:

The molarity of the solution aqueous KOH is 0.25 Molar.

Explanation:

Given,

(w/v) % of aqueous KOH solution = 1.4

To find,

The molarity of aqueous KOH solution (M)

Concept,

Molarity or molar concentration is the number of moles of solute(n) per liter of solution in liters (V).

M = n/V(in litres)

Calculation,

As

weight of solute (g)/volume of the solution (ml)% = 1.4%

⇒ w/v(ml) × 100 = 1.4

⇒ w/v(ml) = 0.014

We know that the molarity (M) is given by:

$M = \frac{weight \ of \ solute}{Molecular \ weight} \times \frac{1000}{Volume \ of \ solution(ml)}$

We substitute the values in the above equation, and we get:

⇒ M = (0.014 × 1000)/G.M.W

⇒ M = 14/56

⇒ M = 1/4 = 0.25 M

Therefore, the molarity of the solution aqueous KOH is 0.25 Molar.

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