Answer:
[tex]M=0.333M[/tex]
Explanation:
Hello,
In this case, the first step is to compute the molar mass of sugar, more specifically sucrose, [tex]C_{12}H_{22}O_{11}[/tex] as shown below:
[tex]M_{C_{12}H_{22}O_{11}}=12g/mol*12+1g/mol*22+16g/mol*11=342g/mol[/tex]
Afterwards, we compute the moles contained into 34.2 grams of sugar, based on the previously computed molar mass as follows:
[tex]n_{C_{12}H_{22}O_{11}}=34.2gC_{12}H_{22}O_{11}*\frac{1molC_{12}H_{22}O_{11}}{342gC_{12}H_{22}O_{11}}=0.1molC_{12}H_{22}O_{11}[/tex]
Finally, we apply the molarity equation taking into account that the 300.0 mL must be used in liters rather than milliliters, that is 0.3 L:
[tex]M=\frac{n_{C_{12}H_{22}O_{11}}}{V_{solution}} = \frac{0.100molC_{12}H_{22}O_{11}}{0.3000L}\\M=0.333M[/tex]
Best regards.
