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A {millnice} mL solution is made by mixing {gramsnice} g of {chem[2]} and water. The percentage weight per volume of the solution is [[0]] % w/v of {chem[2]}.

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Percentage weight per volume is the weight of the solute in grams divided by the amount of solution in millilitres expressed as a percentage. That is, we do the division and multiply by 100.

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In particular, if a {millnice} mL solution is made by mixing {gramsnice} g of {chem[2]} and water. The percentage weight per volume of the solution is:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\dfrac{\\text{grams}}{\\text{millilitres}}\\times 100$$=$$\\dfrac{\\var{gramsnice}}{\\var{millnice}}\\times 100$ % w/v
$=$$\\var{ansnice}$ % w/v 
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A {millb} mL solution is made by mixing {gramsb} g of {chem[3]} and water. The percentage weight per volume of the solution is [[0]] % w/v of {chem[3]}.

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Percentage weight per volume is the weight of the solute in grams divided by the amount of solution in millilitres expressed as a percentage. That is, we do the division and multiply by 100.

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In particular, if a {millb} mL solution is made by mixing {gramsb} g of {chem[3]} and water. The percentage weight per volume of the solution is:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\dfrac{\\text{grams}}{\\text{millilitres}}\\times 100$$=$$\\dfrac{\\var{gramsb}}{\\var{millb}}\\times 100$ % w/v
$=$$\\var{ansb}$ % w/v 
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A {mill} mL solution is made by mixing {grams} g of {chem[0]} and water. The percentage weight per volume of the solution is [[0]] % w/v of {chem[0]} (to 2 decimal places).

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Percentage weight per volume is the weight of the solute in grams divided by the amount of solution in millilitres expressed as a percentage. That is, we do the division and multiply by 100.

\n

\n

In particular, if a {mill} mL solution is made by mixing {grams} g of {chem[0]} and water. The percentage weight per volume of the solution is:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\dfrac{\\text{grams}}{\\text{millilitres}}\\times 100$$=$$\\dfrac{\\var{grams}}{\\var{mill}}\\times 100$ % w/v
$=$$\\var{ans}$ % w/v  (to 2 dec. pl)
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A {per} % {chem[1]} solution contains {amount} g of {chem[1]} in every...

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100 mL.

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200 mL.

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500 mL.

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1000 mL.

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It is important to know {per} % {chem[1]} solution contains {per} g of {chem[1]} in every 100 mL of solution.

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Given this, since we have {amount} g of {chem[1]} we have two times the standard amount and so we must have two times the solution, that is 200 mL.

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Given this, since we have {amount} g of {chem[1]} we have ten times the standard amount and so we must have ten times the solution, that is 1000 mL.

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