Question

Calculate the molarity of 34.2 grams of sugar c12h22o11 in 300.0 ml of solution

Answer 1

`M=0.333M`

Step-by-step explanation:

Hello,

In this case, the first step is to compute the molar mass of sugar, more specifically sucrose,
`C_(12)H_(22)O_(11)` as shown below:

`M_{C_(12)H_(22)O_(11)}=12g/mol*12+1g/mol*22+16g/mol*11=342g/mol`

Afterwards, we compute the moles contained into 34.2 grams of sugar, based on the previously computed molar mass as follows:

`n_{C_(12)H_(22)O_(11)}=34.2gC_(12)H_(22)O_(11)*(1molC_(12)H_(22)O_(11))/(342gC_(12)H_(22)O_(11))=0.1molC_(12)H_(22)O_(11)`

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:

`M=\frac{n_{C_(12)H_(22)O_(11)}}{V_(solution)} = (0.100molC_(12)H_(22)O_(11))/(0.3000L)\\M=0.333M`

Best regards.

Answer 2

molarity = moles /volume in liters

find moles = mass/molar mass

= 34.2g / 342 g/mol= 0.1 moles


molarity is therefore= 0.1/300 x1000 = 0.3 M