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3 years ago

SITUATION # 2: Every morning we drink milk to strengthen our bones. You

Chemistry

1 answer:

hram777 [196]3 years ago

6 0

Answer:

Amount of milk in grams = 95 gram

Explanation:

Given:

Volume of mix solution = 100 ml

Percent of powered milk = 5%

Find:

Amount of milk in grams

Computation:

Assume;

1 ml = 1 gram

Amount of milk = 100 - 100(5%)

Amount of milk = 100 - 5

Amount of milk = 95 ml

Amount of milk in grams = 95 gram

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A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so

Xelga [282]

Answer : The equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

Explanation :

First we have to calculate the initial moles of $Fe^{3+}$ and $SCN^-$ .

$\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}$

$\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol$

and,

$\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}$

$\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol$

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$Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)$

Since 1 mole of $Fe^{3+}$ reacts with 1 mole of $SCN^-$ to give 1 mole of $FeSCN^{2+}$

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$\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M$

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

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A = absorbance of solution

C = concentration of solution

l = path length

$\epsilon$ = molar absorptivity coefficient

$\epsilon$ and l are same for stock solution and dilute solution. So,

$\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}$

For trial solution:

The equilibrium concentration of $SCN^-$ is,

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{initial}$ = 0.00050 M

Now calculate the $[FeSCN^{2+}]$ .

$C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M$

Now calculate the concentration of $SCN^-$ .

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)$

$[SCN^-]_{eqm}=4.58\times 10^{-8}M$

Therefore, the equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

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