Question

Q14. Calculate the volume of
95% alcohol required to
prepare 600 ml of 70%
alcohol.
A. 157.9 ml
B. 211.05 ml
C. 368.p ml
D. 422.1 ml
1
Answer
​

Answer 1

The volume of 95% alcohol is 422.1 ml required to prepare 600 ml of 70% alcohol.

Given :

600 ml of 70% alcohol

To find :

The volume of 95% alcohol is 422.1 ml required to prepare 600 ml of 70% alcohol is

A. 157.9 ml

B. 211.05 ml

C. 368.0 ml

D. 422.1 ml

Solution :

Step 1 of 2 :

Form the equation

Here we have to find the volume of 95% alcohol required to prepare 600 ml of 70% alcohol

c₁ = Concentration of Solution 1 = 95%

c₂ = Concentration of Solution 1 = 70%

v₁ = Volume of Solution 1 = ?

v₂ = Volume of Solution 2 = 600 ml

Using Dilution formula we have

c₁v₁ = c₂v₂

$\displaystyle \sf{ \implies 95 \times v_1 = 70 \times 600 }$

Step 2 of 2 :

Find the required volume

$\displaystyle \sf{ 95 \times v_1 = 70 \times 600 }$

$\displaystyle \sf{ \implies v_1 = \frac{70 \times 600}{95} }$

$\displaystyle \sf{ \implies v_1 = 422.1 }$

Therefore volume of 95% alcohol is 422.1 ml required to prepare 600 ml of 70% alcohol.

Hence the correct option is D. 422.1 ml

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