Q: Calculate the morality of \$NaOH\$ in the solution prepared by dissolving its \$4\ g\$ in water to make solution of \$250\ ml\$.

Weight of \$NaOH =4 g\$ Molecular weight of \$NaOH =40 G\$ Volume of solution is \$250ml\$ Molarity = ? \$M=\frac{W}{MW}\times \frac{1000}{V in ml}\$ \$M=\frac{4}{40}\times \frac{1000}{250}\$\$0.4M\$

Q: Density of a 2.05 M solution of acetic acid in water is 1.02 g/ml. The molality of the solution is:

Suppose the volume of solution is 1000 ml.\$1.02\ gm\$ in 1 ml of soln.\$1020\ gm\$ in 1000 ml of soln.Number of moles \$=2\cdot 05M\times 1\ lit\$\$=2. 05 \ mol\$Mass of solute \$ =n\times M\cdot wt\$\$=2. 05\times 60=123\ gm\$Mass of solvent \$=1020-123=897\ gm\$Molality \$=\dfrac {123}{60}\times \dfrac {1000}{897}\$\$=2.28\ mol \ kg^{-1}\$

Q: Calculate molality of 2.5 grams of Ethanoic Acid \$(CH_3COOH)\$ in 75 grams of Benzene.

The molar mass of ethanoic acid is \$\displaystyle 2(12)+2(16)+4(1)=60\$ g/mol. The mass of ethanoic acid is divided with molar mass of ethanoic acid to obtain number of moles of ethanoic acid.Number of moles of ethanoic acid \$\displaystyle = \dfrac {2.5 \: g}{60 \: g/mol}=0.04167 \: mol\$The mass of benzene is 75 grams or \$\displaystyle \dfrac {75}{1000}=0.075\$ kg.The molality of ethanoic acid in benzene is the ratio of the number of moles of ethanoic acid to the mass of benzene (in kg). It is \$\displaystyle \dfrac {0.04167 \: mol}{0.075 \: kg}=0.556 \: m\$

Q: Equal length of magnesium ribbon are taken in two test tubes \$'A'\$ and \$'B'\$. \$H_2SO_4\$ is added to test tube \$'A'\$ and \$H_2CO_3\$ in the test tube \$'B'\$ in equal amounts. (a) Identify the test tube showing vigorous reaction(b) Give reason to support your answer.(c) Name the gas liberated in both the tubes. How will your prove its liberation?(d) Write chemical equations for both the reaction.(e) Out of the two acids taken above: (i) Which one will have lower pH value?(ii) Lower \$H^+\$ concentration respectively.

a) \$H_2SO_4\$ test tube i.e \$A\$ will show more rigorous reactionsb) Since acidic strength of \$H_2SO_4\$ is more than \$H_2CO_3\$ c) \$Mg+ H_2SO_4 \rightarrow MgSO_4 + H_2 (g)\$ \$Mg+ H_2CO_3 \rightarrow MgCO_3 + H_2(g)\$ The liberated gas is \$H_2\$ and it burns with popping sound. d) \$Mg+ H_2SO_4 \rightarrow MgSO_4 + H_2 (g)\$ \$Mg+ H_2CO_3 \rightarrow MgCO_3 + H_2(g)\$ e) i) Since \$H_2SO_4\$ is more acidic, it will have lower pH value. ii) Test tube \$B\$ will have lower \$H^+\$ concentration as \$H_2CO_3\$ is a weak acid.

Question 1

Calculate the mass of urea $NH_2CONH_2$ required in making 2.5 kg of 0.25 molal aqueous solution.

Accepted Answer

0.25 molal aqueous solution of urea means that 0.25 mole of urea are present in 1 kg of water  Moles of urea = $0.25\, mole$ Mass of solvent (water) = 1 kg = 1000 g Molar mass of urea ($NH_2CONH_2$) = 60 g mol1. $0.25 mole of urea = 0.25\, mol\, \times 60 g\, mol\, 1 = 15 g$ ${\text Total\, mass\, of\, solution = 1000 + 15 g = 1015 g = 1.015 kg}$ Thus; 1.015 kg of solution contain urea = $15 g \times 2.5 kg$ of solution will require urea$\dfrac{15g}{1.015 kg } \times 2.5 kg $ = 37 g

Question 2

Convert into mole.$(a)$ $12$ g of oxygen gas$(b)$ $20$ g of water$(c)$ $22$ g of carbon dioxide.

Accepted Answer

(a) Molecular weight of oxygen in grams = 32 g of oxygen= one mole of oxygen gasThus, 12 g of oxygen$=\frac { 1 }{ 32 } \times 12=0.375\ moles$(b) Molecular weight of water $(H_2O)$ in grams= 2+ 16=18 g=one mole of waterThus, 20 g of water$=\frac { 1 }{ 18 } \times 20=1.111\ moles$(c) Molecular weight of carbon dioxide $(CO_2)$ in grams$= 12+2\times 16=44\ g=$one mole of carbon dioxideThus, 22 g of carbon dioxide $= \frac {1}{44}\times 22=0.5\ moles$

Question 3

The density of a 3M solution of NaCl is 1.25g/ml. Calculate the molality of the solution.

Accepted Answer

3 Molar solution means  there are 3 moles of  NaCl salt in 1 Liter.  Molecular weight of NaCl = 58.44.    Hence, there are $3 \times 58.44$ gms in 1 Litre of water.Density $=\dfrac{ mass}{volume}$Mass of 1 litre of solution = 1.25 gms/ml * 1000 ml = 1, 250 gmsV = volume of water added to make the solution  175.32 gms + 1 gm /ml * V = 1, 250 gm       $\implies V =  1, 074.68 ml  $So 1,074.68 ml or 1, 074.68 gms of water is mixed with 3 moles of Na Cl to make the 3 M solution.            Molality = mass of solute in number of moles / mass of solvent in kg               $= \dfrac{ 3 }{ 1.07468}$kg               $=  2.7915 $ Molal

Question 4

Calculate the morality of $NaOH$ in the solution prepared by dissolving its $4\ g$ in water to make solution of $250\ ml$.

Accepted Answer

&#10;&#10;Weight of $NaOH =4 g$&#10;&#10;Molecular weight of $NaOH =40 G$&#10;&#10;Volume of solution is $250ml$&#10;&#10;Molarity = ?&#10;&#10;$M=\frac{W}{MW}\times \frac{1000}{V in ml}$&#10;&#10;$M=\frac{4}{40}\times \frac{1000}{250}$$0.4M$&#10;&#10;&#10;&#10;

Question 5

Density of a 2.05 M solution of acetic acid in water is 1.02 g/ml. The molality of the solution is:

Accepted Answer

Suppose the volume of solution is 1000 ml.$1.02\ gm$ in 1 ml of soln.$1020\ gm$ in 1000 ml of soln.Number of moles $=2\cdot 05M\times 1\ lit$$=2. 05 \ mol$Mass of solute $ =n\times M\cdot wt$$=2. 05\times 60=123\ gm$Mass of solvent $=1020-123=897\ gm$Molality $=\dfrac {123}{60}\times \dfrac {1000}{897}$$=2.28\ mol \ kg^{-1}$

Question 6

Calculate molality of 2.5 grams of Ethanoic Acid $(CH_3COOH)$ in 75 grams of Benzene.

Accepted Answer

The molar mass of ethanoic acid is $\displaystyle 2(12)+2(16)+4(1)=60$ g/mol. The mass of ethanoic acid is divided with molar mass of ethanoic acid to obtain number of moles of ethanoic acid.Number of moles of ethanoic acid $\displaystyle = \dfrac {2.5 \: g}{60 \: g/mol}=0.04167 \: mol$The mass of benzene is 75 grams or $\displaystyle \dfrac {75}{1000}=0.075$ kg.The molality of ethanoic acid in benzene is the ratio of the number of moles of ethanoic acid to the mass of benzene (in kg). It is $\displaystyle \dfrac {0.04167 \: mol}{0.075 \: kg}=0.556 \: m$

Question 7

Determine the empirical formula of an oxide of iron which has $69.9\%$ iron and $30.1\%$ dioxygen by mass.

Accepted Answer

The iron oxide has 69.9% iron and 30.1% dioxygen by mass.Thus, 100 g of iron oxide contains 69.9 g iron and 30.1 g dioxygen.The number of moles of iron present in 100 g of iron oxide are $\displaystyle \frac {69.9}{55.8} = 1.25$.The number of moles of dioxygen present in 100 g of iron oxide are $\displaystyle \frac {30.1}{32} = 0.94$.The ratio of the number of oxygen atoms to the number of iron atoms present in one formula unit of iron oxide is $\displaystyle \frac {2 \times 0.94}{1.25} = 1.5:1=3:2$.Hence, the formula of the iron oxide is $\displaystyle Fe_2O_3 $.

Question 8

A solution contains 40 g of common salt in 320 g of water. Calculate the concentration in terms of mass by mass percentage of the solution.

Accepted Answer

Mas  of common salt (solute)=40 gMass of water (soluted) =320 gmass of solution=$320+40$               =360 gconcentration of solution=$\dfrac{M_{mass} of solute}{mass of solution}\times100$$\dfrac{40}{360}\times 100=11.11%$Hence, the concentration of the solution is 11.11%

Question 9

40 grams of sodium hydroxide is dissolved in enough water to make $1$ liter of solution. Determine the molarity of the solution?

Accepted Answer

The molar mass of NaOH is 40 g/mol.The number of moles of NaOH is equal to the ratio of its mass to molar mass.The number of moles of NaOH $\displaystyle  = \dfrac {40 g}{40 g/mol} = 1 mol$ The molarity of NaOH solution is the ratio of number of moles of NaOH to total volume of solution in L.$\displaystyle  M = \dfrac {1mol}{1L}=1 M$

Question 10

Equal length of magnesium ribbon are taken in two test tubes $'A'$ and $'B'$. $H_2SO_4$ is added to test tube $'A'$ and $H_2CO_3$ in the test tube $'B'$ in equal amounts. (a) Identify the test tube showing vigorous reaction(b) Give reason to support your answer.(c) Name the gas liberated in both the tubes. How will your prove its liberation?(d) Write chemical equations for both the reaction.(e) Out of the two acids taken above: (i) Which one will have lower pH value?(ii) Lower $H^+$ concentration respectively.

Accepted Answer

a) $H_2SO_4$ test tube i.e&#10;$A$ will show more rigorous reactionsb) Since acidic strength of&#10;$H_2SO_4$ is more than $H_2CO_3$&#10;&#10;c) $Mg+ H_2SO_4 \rightarrow&#10;MgSO_4 + H_2 (g)$&#10;&#10;    $Mg+ H_2CO_3 \rightarrow&#10;MgCO_3 + H_2(g)$&#10;&#10;The liberated gas is $H_2$&#10;and it burns with popping sound.&#10;&#10;d) $Mg+ H_2SO_4 \rightarrow&#10;MgSO_4 + H_2 (g)$&#10;&#10;    $Mg+ H_2CO_3 \rightarrow&#10;MgCO_3 + H_2(g)$&#10;&#10;e) i) Since $H_2SO_4$ is&#10;more acidic, it will have lower pH value. &#10;&#10;ii) Test tube $B$ will have lower $H^+$ concentration as&#10;$H_2CO_3$ is a weak acid.