Question

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Answer 1

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AnsWer-1)

• density of HNO3 = 1.41 g/ml
• mass percent of HNO3 = 69%
• Molecular weight of HNO3 = 63g/mol

✒ 69% w/w indicates,

• mass of solute(HNO3) = 69g
• mass of solution = 100g

✒ Volume of solution = mass/density = 70.92 ml

✴ Concentration of nitric acid is given by

$\sf[HNO_3]=\dfrac{W_{HNO_3}\times 1000}{Mw_{HNO_3}\times Vml_{solution}}\\ \\ \sf[HNO_3]=\dfrac{69\times 1000}{63\times 70.92}\\ \\ \boxed{\sf{\pink{[HNO_3]=15.44\:M}}}$

▪ Concentration of nitric acid = 15.44M

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AnsWer-2)

• density of methanol= 0.793kg/L
• volume of solution = 2.5L
• Molarity of solution = 0.25M
• Molecular weight of methanol = 32g/mol

✒ 0.793kg/L = 0.793g/ml

✴ Concentration of methanol is given by

$\sf[Methanol]=\dfrac{W_{Methanol}}{Mw_{Methanol}\times {V_L}_{(solution)}}\\ \\ \sf 0.25=\dfrac{W_{Methanol}}{32\times 2.5}\\ \\ \sf\:Weight\:of\:Methanol=20g$

✒ Volume = weight/density = 20/0.793

▪ Volume of methanol requires to make a solution of concentration 0.25M = 25.22ml

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AnsWer-3)

• weight of sugar = 20g
• Volume of solution = 2L
• Molecular weight of sugar = 342g/mol

✴ Concentration of sugar is given by

$\sf[Sugar]=\dfrac{W_{Sugar}}{Mw_{Sugar}\times {V_L}_{(solution)}}\\ \\ \sf[Sugar]=\dfrac{20}{342\times 2}\\ \\ \boxed{\sf{\orange{[Sugar]=0.0293\:MolL^{-1}}}}$

▪ Concentration of sugar = 0.0293M

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Answer 2

Answer:

density of HNO3 = 1.41 g/ml

mass percent of HNO3 = 69%

Molecular weight of HNO3 = 63g/mol

✒ 69% w/w indicates,

mass of solute(HNO3) = 69g

mass of solution = 100g

✒ Volume of solution = mass/density = 70.92 ml

✴ Concentration of nitric acid is given by

$$\begin{lgathered}\sf[HNO_3]=\dfrac{W_{HNO_3}\times 1000}{Mw_{HNO_3}\times Vml_{solution}}\\ \\ \sf[HNO_3]=\dfrac{69\times 1000}{63\times 70.92}\\ \\ \boxed{\sf{\pink{[HNO_3]=15.44\:M}}}\end{lgathered}$$

▪ Concentration of nitric acid = 15.44M

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AnsWer-2)

density of methanol= 0.793kg/L

volume of solution = 2.5L

Molarity of solution = 0.25M

Molecular weight of methanol = 32g/mol

✒ 0.793kg/L = 0.793g/ml

✴ Concentration of methanol is given by

$$\begin{lgathered}\sf[Methanol]=\dfrac{W_{Methanol}}{Mw_{Methanol}\times {V_L}_{(solution)}}\\ \\ \sf 0.25=\dfrac{W_{Methanol}}{32\times 2.5}\\ \\ \sf\:Weight\:of\:Methanol=20g\end{lgathered}$$

✒ Volume = weight/density = 20/0.793

▪ Volume of methanol requires to make a solution of concentration 0.25M = 25.22ml

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AnsWer-3)

weight of sugar = 20g

Volume of solution = 2L

Molecular weight of sugar = 342g/mol

✴ Concentration of sugar is given by

$$\begin{lgathered}\sf[Sugar]=\dfrac{W_{Sugar}}{Mw_{Sugar}\times {V_L}_{(solution)}}\\ \\ \sf[Sugar]=\dfrac{20}{342\times 2}\\ \\ \boxed{\sf{\orange{[Sugar]=0.0293\:MolL^{-1}}}}\end{lgathered}$$

▪ Concentration of sugar = 0.0293M