Question

a chemist has one solution is 50%acid and another solution which is 25%acid.how much each should be mixed to make 10litres of 40%acid solution?(use cramers rule)​

Answer 1

Let the amount of 50% acid be x liters.

And the amount of 25% acid be y litres.

To make 10 litres of 40% acid solution,

\begin{lgathered}\begin{array} { c } { 0.5 x + 0.25 y = 4 \ldots \ldots ( i ) } \\\\ { x + y = 10 \ldots \ldots . ( i i ) } \end{array}\end{lgathered}

0.5x+0.25y=4……(i)

x+y=10…….(ii)

[balancing concentrations]

Multiply the equation (i) by 10.

Then the equation (i) becomes,

( i ) \times 10 \rightarrow 5 x + 2.5 y = 40 \rightarrow ( i i i )(i)×10→5x+2.5y=40→(iii)

And multiply the equation (ii) by 5

Then the equation (ii) becomes,

( i i ) \times 5 \rightarrow 5 x + 5 y = 50 \rightarrow ( i v )(ii)×5→5x+5y=50→(iv)

By subtracting, equation (iii) from (iv)

Subtracting,

\begin{lgathered}\begin{array} { c } { 5 x + 2.5 y = 40 } \\\\ { 5 x + 5 y = 50 } \\\\ { - 2.5 y = - 10 } \end{array}\end{lgathered}

5x+2.5y=40

5x+5y=50

−2.5y=−10

Then,

y = \frac { 10 } { 2.5 } = 4y=

2.5

10

=4

Substituting the value of y, (y = 4) in equation (iii) we get,

\begin{lgathered}\begin{array} { c } { 5 x + 2.5 y = 40 } \\\\ { 5 x + 2.5 ( 4 ) = 40 } \\\\ { 5 x + 10 = 40 } \\\\ { 5 x = 30 } \\\\ { x = 6 } \end{array}\end{lgathered}

5x+2.5y=40

5x+2.5(4)=40

5x+10=40

5x=30

x=6

Hence, 6 litres of 50% solution and 4 litres of 25% solution are needed to form 10 litres of 40% of acid solutions.

Answer 2

plz refer to this attachment