Question

plz help me in these three questions ​

Answer 1

4.)

2HCl + Na2CO3--->H2CO3+ 2NaCl

2V 1V

2mol 1mol

36.5g HCl in 1000mL=1 molar HCl

So,

3.65g HCl in 100mL=1 molar HCl Solution is given.

105.99g Na2CO3 in 1000mL= 1 molar Na2CO3.

So,

10g Na2CO3 in 1000mL=0.094 molar Na2CO3.

Thus,

10g of Na2CO3 in 100 mL=0.94 molar Na2CO3.

Now from the chemical equation, 100mL of 1 molar HCl requires 100mL 0.5 molar Na2CO3

So, 0.94 M Na2CO3

100×0.5= Y × 0.94

Therefore Y = 53.19 mL Na2CO3

5.)

15% solution

Mass = volume× density

volume=100 cm3

density=1.10 g/cm3

mass of 15% solution = 100×1.10=110 g

mass of sulphuric acid in it =.15×110=16.5g

mass of water present in it =

110-16.5=93.5 g of water

95% solution

let mass of 95% solution required to prepare 15% H2SO4 with 16.5 g acid in it be x

16.5g =0.95× x

x=16.5×100/95 =17.4 g

Volume of 95%H2SO4 needed to prepare 15%H2SO4

VOLUME =mass/density

Volume =17.4/1.85= 9.4 cm3

Volume water needed to make up the volume =100-9.4=90.6 cm3

6.)

95% of ethanol by amt solution consists of 95 g of ethanol is 5 g of water

95g of ethanol 95/45 mole

molecular wt of ethanol = 45g

5g of water = 5/18 mole

Xethanol=(methanol)÷(methanol+nwater)

=(95/45)÷{(95/45)+(5/18) }

Xethanol=2.111/(2.111+0.278 )= 0.881

Xwater=1-0.881=0.119

Hope you got my answer.