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Standard X

Mathematics

Relationship between LCM and GCD

For preparing...

Question

For preparing $3.00 L$ of $1 M N a O H$ by mixing portions of two stock solutions ( $A$ and $B$ ) of $2.50 M N a O H$ and $0.40 M N a O H$ respectively. Find out the amount of $B$ stock solution (in L) added :

8.57 L

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2.14 L

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1.28 L

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7.51 L

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Solution

The correct option is D $2.14 L$
Let $x$ L of $B$ stock solution is added.
$3.00 - x$ L of $A$ stock solution will be added.
$M V = M_{A} V_{A} + M_{B} V_{B}$
$1.00 M \times 3.00 L = 2.50 M \times (3.00-x) L + 0.40 M \times x L$
$3.00 = 7.50 - 2.50 x + 0.40 x$
$4.50 = 2.10 x$
$x = 2.14$ L

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For preparing 3.00 L of 1M NaOH by mixing portions of two stock solutions (A and B) of 2.50 M NaOH and 0.40 M NaOH respectively. Find out the amount of B stock solution (in L) added :

The correct option is D 2.14 LLet x L of B stock solution is added.3.00−x L of A stock solution will be added.MV=MAVA+MBVB1.00 M × 3.00 L = 2.50 M × (3.00-x) L+ 0.40 M × x L3.00=7.50−2.50x+0.40x4.50=2.10xx=2.14 L

For preparing $3.00 L$ of $1 M N a O H$ by mixing portions of two stock solutions ( $A$ and $B$ ) of $2.50 M N a O H$ and $0.40 M N a O H$ respectively. Find out the amount of $B$ stock solution (in L) added :

The correct option is D $2.14 L$
Let $x$ L of $B$ stock solution is added.
$3.00 - x$ L of $A$ stock solution will be added.
$M V = M_{A} V_{A} + M_{B} V_{B}$
$1.00 M \times 3.00 L = 2.50 M \times (3.00-x) L + 0.40 M \times x L$
$3.00 = 7.50 - 2.50 x + 0.40 x$
$4.50 = 2.10 x$
$x = 2.14$ L