Question

HEYA MATES

HERE IS A NEW QUESTION FOR YOU ALL.

✏ I REFER TO THE ABOVE PIC FOR ANSWER

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✒ ✏ QUESTIONS 9,10 AND 11. ✒ ✏

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Answer 1

Q9. Let the volume of 6M HCl be x Litres then volume of 2 M HCl is (1-x) Litres.
M1V1 + M2V2 = M3V3
=> 6x + 2(1-x) = 3×2
=> 6x +2 -2x = 6
=> 4x = 4
=> x = 1
and (1-x) = 0
Hence, volume of 6 M HCl required is 1 L
and volume of 2 M HCl required is 0 L

Hope it helps!!!

Answer 2

09)

We know that :-

$\boxed{\bold{\sf{\red{Molarity= \dfrac{No.\:of\:moles\:of\:solute}{Volume\:of\:solution(in\:L)}}}}}$

So,

Let the volume of 6M HCl be v1 and volume of 2M HCl be v2.

When mixed,

It produces 2 Litres of 3M HCl solution.

Now,

According to Molarity of Mixtures, we know that :-

$\boxed{\bold{\sf{M_3 =\dfrac{M_1V_1+M_2V_2}{V_1+V_2}}}}$

Thus,

According to the above equation we have :-

$3=\dfrac{6V_1+2V_2}{2}\\ \\ \\6V_1 +2V_2=6\\ \\ \\6V_1+2(2-V_1)=6\\ \\ \\6V_1+4-2V_1=6\\ \\ \\4V_1=2\\ \\ \\V_1=0.5$

Hence,

Volume of 6M HCl required is 0.5 L

And,

Volume of 2M HCl required is 2-0.5 = 1.5Litres

$\rule{200}{2}$

10)

Mass of Calcium Carbonate = 20 Gram

Mass of Hydrochloric Acid = 20 gram

According to Reaction :-

$Mole\:Ratio=\dfrac{moles\:of\:element/compound}{stoichiometric\:coefficients}$

The mole ratio of Calcium carbonate is 0.2 while the mole ratio of Hydrochloric Acid is 0.27.

Thus,

The Limiting Reagent in the reaction is Calcium Carbonate.

1 mole of CaCO3 reacts to produce 1 Mole of Carbondioxide.

Moles of CaCO3 given is :-

$= \frac{20}{100} \\ \\ \\ = 0.2 \: moles$

Hence,

0.2 moles of CO2 will be produced :-

Mass of CO2 produced will be :-

$= 0.2 \times 44 \\ \\ \\ = 8.8 \: grams$

$\rule{200}{2}$

11)

H2SO4 is labeled as 98% by mass

This means that 98g of Sulfuric Acid is present in 100g of solution.

Number of moles of H2SO4 will be :-

$= \frac{98}{98} \\ \\ \\ = 1 \: mole$

Density given = 1.84 g/cm³

Thus,

Density = mass/volume

Volume = mass/density

$v = \frac{100}{1.84} \\ \\v = 54.34 {cm}^{3} \\ \\ \\v = 54.34 \: ml$

Now,

Molarity = moles/volume(in L)

$M=\dfrac{1*1000}{54.34}=18.4M$

According to Neutralization law of Molarity :-

$M_1V_1=M_2V_2$

Hence ,

$18.4 \times v1 = 0.5 \times 5 \\ \\ \\v1 = \frac{2.5}{18.4} \\ \\ \\v1 = 0.1358l \\ \\ \\v1 =135.8ml$

$\rule{200}{2}$