Chemistry - Concentration Terms Question with Solution | TestHub

$d_{solution}=1.071g/mL$.

$X_{solute}=\frac{1}{101}$⇒ 1 mol solute in 100 mol solvent.

$W_{sugar}=342g$.

&$W_{H_{2}O}=100\times 18=1800g$.

$W_{solution}=1800+342=2142g$.

$V_{solution}=\frac{W_{solution}}{d_{solution}}=\frac{2142g}{1.071g/mL}=2000mL=2L$

Therefore, the molarity of the solution is:

$Molarity=\frac{molesofsolute}{volumeofsolution(L)}=\frac{1mol}{2L}=0.5M$