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Paul [167]
3 years ago
11

If 255 g of water has 10.0 g of NaCl dissolved into initially, how much NaCl must be added in order to raise the mass percent of

NaCl by 10%?
Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
8 0
M(H₂O) = 255 g.
m₁(NaCl) = 10.0 g.
m₁(solution) = 255 g + 10 g = 265 g.
ω₁ = 10 g / 265 g · 100%.
ω₁ = 3.77% ÷ 100% = 0.0377.
ω₂= 10% ÷ 100% = 0.1.
ω₂= m₁(NaCl) + m₂(NaCl) / m₁(solution) + m₂(NaCl).
0.1 = 10 g + m₂(NaCl) / 265 g + m₂(NaCl).
26.5 g + 0.1·m₂(NaCl) = 10 g + m₂(NaCl).
0.9·m₂(NaCl) = 16.5 g.
m₂(NaCl) = 18.33 g.
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This is an incomplete question, here is a complete question.

Suppose we now collect hydrogen gas, H₂(g), over water at 21°C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:

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Explanation :

First we have to calculate the number of moles of aluminium.

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Molar mass of aluminium = 27 g/mol

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

\text{Moles of aluminium}=\frac{1.35g}{27g/mol}=0.05mol

The given chemical reaction is:

2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)

As, hydrochloric acid is present in excess. So, it is considered as an excess reagent.

Thus, aluminium is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of hydrogen gas

So, 0.005 moles of aluminium will produce = \frac{3}{2}\times 0.05=0.0750mol of hydrogen gas

Now we have to calculate the mass of helium gas by using ideal gas equation.

PV = nRT

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V = Volume of the helium gas = ?

n = number of moles of hydrogen gas = 0.075 mol

R = Gas constant = 62.364\text{ L Torr }mol^{-1}K^{-1}

T = Temperature of hydrogen gas = 21^oC=[21+273]K=294K

Now put all the given values in above equation, we get:

743Torr\times V=0.075mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 294K\\\\V=1.85L

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