Four students give their teacher identical apples. Each student sets his or her apple on a different stack of books. Which

Chemistry

1 answer:

ycow [4]4 years ago

7 0

D
explanation ; i’m smart

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What is per cent by mass of Na2CO3 in a 1.0 molal aqueous solution ?

aivan3 [116]

Hello!

We have the following data:

MM (Molar mass of Na2CO3)
Na = 2*23 = 46 u
C = 1*12 = 12 u
O = 3*16 = 48 u
----------------------
MM (Molar mass of Na2CO3) = 46+12+48 = 106 g/mol
n (mol number) = 1 mol
m1 (mass of the solute) = ?

$n = \dfrac{m_1}{MM}$

$1\:mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}= \dfrac{m}{106\:g/mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}}$

$m_1 = 1*106$

$\boxed{m_1 = 106\:g}\Longleftarrow(solute\:mass)$

So if we have the dissolution in a liter (m2 - mass of solvent), that is 1000g , then the total mass of the solution (m) will be:

$m\:(solution\:mass) = m_1\:(solute\:mass) + m_2\:(solvent\:mass)$

$m = 106\:g+1000\:g$

$\boxed{m = 1106\:g}\Longleftarrow(solution\:mass)$

<span>Now, let's find the percentage in mass (% m / m), let's see:
</span>
$\%(m/m) = \dfrac{m_1}{m}*100$

$\%(m/m) = \dfrac{106}{1106}*100$

$\%(m/m) = \dfrac{10600}{1106}$

$\%(m/m) = 9.58408...\to\:\boxed{\boxed{\%(m/m) \approx 9.6}}\end{array}}\qquad\checkmark$

Answer: B) 9.6

I Hope this helps, greetings ... DexteR!

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