Question

How many grams of solute are present in $50{\text{ }}mL$ of $0.360{\text{ }}M$ sodium chloride?

Answer 1

The molality of an answer is characterized as the quantity of moles of solute in an answer isolated by the heaviness of solvent in kilograms. Molarity is a concentration term.
$concentration = \dfrac{{Moles{\text{ }}of{\text{ }}solute}}{{Volume{\text{ }}of{\text{ }}solution}}$

Complete step by step answer: In this way, in the inquiry, it is given that an answer of sodium chloride is available with a volume of $50mL$ and the arrangement has a molarity of$0.360M$. Furthermore, from the data given in the inquiry we need to discover the moles of solute. Thus, to address the inquiry, we examine some fundamental ideas of arrangement, its parts and different boundaries of concentration of the arrangement. We know about the term arrangement from the time we are considering science and we should catch up on certain ideas about arrangements.
How about we start off with the condition for molarity:
$Molarity = \dfrac{{Moles{\text{ }}of{\text{ }}solute}}{{Liters{\text{ }}of{\text{ }}solution}}$
We are given the molarity and the volume of arrangement. The solitary issue is that the volume is given in $mL$ rather than $L$.
This issue can be fixed by utilizing the accompanying change factor: $1000mL = 1L$
In this way, on the off chance that we partition $50mL$ by $1000mL$ we will acquire an estimation of$0.05L$.
Next, the condition must be modified to address for the moles of solute:
$Moles{\text{ }}of{\text{ }}solute{\text{ }} = {\text{ }}Molarity{\text{ }} \times {\text{ }}Liters{\text{ }}of{\text{ Solution}}$
Presently, multiply $0.360{\text{ }}M$ by $0.05$:
$0.360{\text{ }}\dfrac{{mol}}{{1L}} \times 0.05{\text{ }}L = 0.018{\text{ }}mol$
To get the mass of solute, we should have the molar mass of $NaCl$, which is$58.44{\text{ }}g/mol$.
At long last, increase the quantity of moles by $58.44{\text{ }}g/mol$.
$0.018mol \times 58.44\dfrac{g}{{1mol}} = {\text{ }}1.05g$.
There are $1.05g$ of solute present.

Note:
To change $mL$ over to $L$ for example from a more modest amount to a greater amount the division activity ought to be completed.
$1mL = 1000L = {10^3}L$