The density of water is 1 g-ml- The number of molecules present in 1 litre of water is-18times 6-023times 10-23-1-8times 6-023times 10-23-6-023times 10-23-55-55times 6-023times 10-23-

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Toppr Admin · Accepted Answer

1 g-x2192- 1 ml of water-x2234-1000g -x2192- 1 L of solution-No- of moles -100018-55-55 moles-xA0-So- number of molecules - No- of moles -xD7-6-023-xD7-1023-55-55-xD7-6-023-xD7-1023Hence- option D is correct-

The density of water is $1$ g/ml. The number of molecules present in 1 litre of water is:
$18 \times 6.023 \times 10^{23}$
$1.8 \times 6.023 \times 10^{23}$
$6.023 \times 10^{23}$
$55.55 \times 6.023 \times 10^{23}$

$1$ g $\to$ $1$ ml of water

$∴ 1000$ g $\to$ $1$ L of solution.

No. of moles $= \frac{1000}{18} = 55.55$ moles

So, number of molecules $=$ No. of moles $\times 6.023 \times 10^{23}$

$= 55.55 \times 6.023 \times 10^{23}$

Hence, option D is correct.

The density of water is 1 g/ml. The number of molecules present in 1 litre of water is:18×6.023×10231.8×6.023×10236.023×102355.55×6.023×1023

1 g→ 1 ml of water∴1000g → 1 L of solution.No. of moles =100018=55.55 moles So, number of molecules = No. of moles ×6.023×1023=55.55×6.023×1023Hence, option D is correct.

The density of water is $1$ g/ml. The number of molecules present in 1 litre of water is:
$18 \times 6.023 \times 10^{23}$
$1.8 \times 6.023 \times 10^{23}$
$6.023 \times 10^{23}$
$55.55 \times 6.023 \times 10^{23}$

$1$ g $\to$ $1$ ml of water

$∴ 1000$ g $\to$ $1$ L of solution.

No. of moles $= \frac{1000}{18} = 55.55$ moles

So, number of molecules $=$ No. of moles $\times 6.023 \times 10^{23}$

$= 55.55 \times 6.023 \times 10^{23}$

Hence, option D is correct.