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2 years ago

If u mix 10ml of 70%ethanol with 20ml of 80% ethanol with 50ml water what is the concentration of the resulted solution

Chemistry

1 answer:

morpeh [17]2 years ago

6 0

Answer:

29% is the final concentration

Explanation:

To solve this question we must find the volume of ethanol added and the volume of the whole solution. The concentration will be:

Volume ethanol / Total volume * 100

Volume ethanol:

10mL * 70% = 7mL ethanol

20mL * 80% = 16mL ethanol

= 23mL ethanol

Total volume:

10mL + 20mL + 50mL = 80mL

Concentration:

23mL / 80mL * 100

= 29% is the final concentration

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mote1985 [20]

Answer : The correct expression is, (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Explanation :

First we have to calculate the moles of $Zn(ClO_3)_2$

$\text{ Moles of }Zn(ClO_3)_2=\frac{\text{ Mass of }Zn(ClO_3)_2}{\text{ Molar mass of }Zn(ClO_3)_2}=\frac{150g}{232.29g/mole}=\frac{150}{232.29}mole$

Now we have to calculate the moles of $O_2$

The balanced chemical reaction is,

$Zn(ClO_3)_2\rightarrow ZnCl_2+3O_2$

From the balanced chemical reaction we conclude that,

As, 1 mole of $Zn(ClO_3)_2$ react to give 3 mole of $O_2$

So, $\frac{150}{232.29}$ moles of $Zn(ClO_3)_2$ react to give $\frac{150}{232.29}\times \frac{3}{1}$ moles of $O_2$

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(\frac{150}{232.29}\times \frac{3}{1})mole\times (31.998g/mole)=\frac{150\times 3\times 31.998}{232.29\times 1}g$

Therefore, the correct expression for the mass of oxygen gas formed is, (150 × 3 × 31.998) ÷ (232.29 × 1) grams

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