Question

Alguém me ajuda por favor ??​

Answer 1

$m_{NaCl} = 0,9g + 10,8g = 11,7g\\V_{H_2O} = 100mL + 200mL = 300mL\\\\C = \frac{m_{NaCl}}{V_{H_2O}} = \frac{11,7g}{300mL} = 0.039g/mL\\\\1mL = 0,001l\\\\C = \frac{0,039g}{0,001L} = 39g/L\\\\D. \ 39g/L$