Question #777f6

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Question #777f6

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Answer 1 · 2018-01-10T20:27:52

$= 204.75 g$ $=204.75g$

Explanation:

Given, the required $M$ $M$ and the needed volume for the solution, the mass of the sodium chloride $(N a C l)$ $(NaCl)$ can be computed as;

$Molarity (M) = \frac{no. of moles (η)}{Li Solution (V)}$ $"Molarity"(M)=("no. of moles"(eta))/("Li Solution"(V))$

where:

$η = \frac{m a s s (m)}{Molar mass (M m)}$ $eta=(mass(m))/("Molar mass"(Mm)$

Now find the molar mass of $N a C l$ $NaCl$ . Value is obtainable from the periodic table; i.e.,
$N a C l = \frac{58.5 g}{m o l}$ $NaCl=(58.5g)/(mol)$

Substitute the value of $η$ $eta$ to the original formula. Rearrange it to isolate the $m$ $m$ and plug in given values to solve the required data as shown below;

$M = \frac{\frac{m}{M m}}{V}$ $M=(m/(Mm))/V$

$\frac{m}{M m} = M V$ $(m)/(Mm)=MV$

$m = M \times V \times M m$ $m=MxxVxxMm$

$m=(3.5cancel(mol))/(cancel(L))xx(1cancel(L))xx(58.5g)/(1cancel(mol))$

$m=204.75g$

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Question #777f6

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