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What is the concentration in #"%w/v"# of sodium sulfate when #"9.74 g"# of it is placed into a #165# volume?

Answer 1

Peyton Adams

Well... #165# what? I assume #"mL"# (because who would ever drop #"9.74 g"# of something in a huge #"165 L"# vat?)...

And thus, with no other assumptions...

#c_(Na_2SO_4) = ("9.74 g Na"_2"SO"_4)/("165 mL soln")#

#%w//v# is defined for a solute mass in #"g"# and a solution volume in #"mL"#, so we are basically done...

#color(blue)(c_(Na_2SO_4)) = 0.0590 xx 100% = color(blue)(5.90% w//v)#

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Answer 2

Noah Aldrich

To find the concentration of sodium sulfate in %w/v:

[ \text{%w/v} = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100 ]

Substitute the given values:

[ \text{%w/v} = \frac{9.74 , \text{g}}{165 , \text{mL}} \times 100 ]

[ \text{%w/v} = \frac{9.74}{165} \times 100 ]

[ \text{%w/v} \approx 5.90% ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.