JEE MAIN - Chemistry (2021 - 24th February Morning Shift - No. 19)
4.5 g of compound A (MW = 90) was used to make 250 mL of its aqueous solution. The
molarity of the solution in M is x $$ \times $$ 10-1.
The value of x is _______. (Rounded off to the nearest integer)
The value of x is _______. (Rounded off to the nearest integer)
Answer
2
Explanation
Given, weight of compound A = 4.5 g
Molecular weight of compound A = 90 g/mol
Volume of solution (in mL) = 250 mL
Now, molarity is defined as number of moles of solute or compound A divided by volume of solution (in L).
$$M = {{Number\,of\,moles\,of\,solute\,(n)} \over {Volume\,of\,solution}}$$
$$ = {{{{4.5} \over {90}}} \over {{{250} \over {1000}}}} = 0.2$$ = 2 $$\times$$ 10$$-$$1 M
$$\therefore$$ $$n = {{Weight\,of\,solute\,(compound\,A)} \over {Molecular\,weight\,of\,solute\,(compound\,A)}}$$
Hence, x $$\times$$ 10$$-$$1 $$\mu$$
x = 2
Molecular weight of compound A = 90 g/mol
Volume of solution (in mL) = 250 mL
Now, molarity is defined as number of moles of solute or compound A divided by volume of solution (in L).
$$M = {{Number\,of\,moles\,of\,solute\,(n)} \over {Volume\,of\,solution}}$$
$$ = {{{{4.5} \over {90}}} \over {{{250} \over {1000}}}} = 0.2$$ = 2 $$\times$$ 10$$-$$1 M
$$\therefore$$ $$n = {{Weight\,of\,solute\,(compound\,A)} \over {Molecular\,weight\,of\,solute\,(compound\,A)}}$$
Hence, x $$\times$$ 10$$-$$1 $$\mu$$
x = 2
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