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1.the molality of dilute aqua solution of 0.05 M NaOH is
options:-
a)0.05 m
b)18.5 m
c)55.55 m
d)0.1 m
_________________________________________
2. The number of molecules present in One drop(0.05ml) of benzene(density = 0.78 g/ml)is nearly equal to![{N_a = 6 \times {10}^{23}} {N_a = 6 \times {10}^{23}}](https://tex.z-dn.net/?f=%7BN_a+%3D+6+%5Ctimes++%7B10%7D%5E%7B23%7D%7D)
options:-
a)6 × 10²³
b)3 × 10²¹
c)3 ×.10²⁰
d)6 × 10¹⁹
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Answers
Answered by
129
Solution - 1).
Given:
=> Molarity = 0.05 M
=> Volume of solution = 1 L
To Find:
=> Molality
Formula used:
Here,
No. of moles of solute = 0.05 M
Now, we have to find mass of solution.
Let density of water = 1g/cm³
Then, mass of water = 1Kg
Put the following values in the formula,
Solution - 2).
Given:
=> Volume = 0.05 ml
=> Density = 0.78 g/ml
To Find:
=> The number of molecules present in One drop.
So,
As we know,
So, Mass of 1 drop of benzene = 0.05 × 0.78
= 0.039 g
It is given that 1 mole of benzene contains
Answered by
98
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