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Question #cbe8e

1 Answer

Jan 22, 2016

$M_{1} = 2.5 M$ $M_1=2.5M$

Explanation:

Since the number of mole remains the same, we can then say:

$M_{1} V_{1} = M_{2} V_{2}$ $M_1V_1=M_2V_2$

Where, $M_{1} = ?$ $M_1=?$ , $V_{1} = 10 m L$ $V_1=10mL$ , $M_{2} = 0.1 M$ $M_2=0.1M$ and $V_{2} = 250 m L$ $V_2=250mL$ .

The concentration of the stock solution is then found by:

$=>M_1=(M_2V_2)/(V_1)=(0.1Mxx250cancel(mL))/(10 cancel(mL))=2.5M$

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