Question

Se requiere preparar 1500 ml de un solución de sacarosa C12H22O11 0.3 M ¿Cuántos gramos de sacarosa se deben pesar?

Answer 1

Respuesta:

Se deben pesar 154.0 g de $C_{12}H_{22}O_{11}$.

Explicación:

$(1500\ mL\ de\ C_{12}H_{22}O_{11})(\frac{1\ L}{1000\ mL })(\frac{0.3\ mol\ de\ C_{12}H_{22}O_{11}}{1\ L\ de\ disolucion})(\frac{342.3\ g\ de\ C_{12}H_{22}O_{11}}{1\ mol\ de\ C_{12}H_{22}O_{11}})$$= 154.0\ g\ de\ C_{12}H_{22}O_{11}$