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kotegsom [21]
3 years ago
14

How much calcium hydroxide in grams is needed to produce 1.5 l of a 0.25m solution?

Chemistry
1 answer:
Vesna [10]3 years ago
6 0
The  mass    ca(Oh)2  needed  to  produce  1.5 l  of  0.25 solution  is  calculated  as follows
find  moles
 
=moles   = molarity   x  volume

moles is therefore =   1.5  x 0.25  =  0.375  moles

mass  =  moles  x  molar  mass
0.375mol  x  74.09 g/mol  =  27.78  grams
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Answer:

0.175\; \rm mol \cdot L^{-1}.

Explanation:

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\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to  \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}.

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Net ionic equation:

\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}.

Calculate the initial quantity of nitrate ions in the mixture.

\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}.

Since nitrate ions \rm {NO_3}^{-} do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol.

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be (1/2) of the concentration in the original solution.

\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}.

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