**Problem Statement:**
Steam at 100 degree Celsius is passed into 22 grams of water at 20 degree Celsius. When the resultant temperature is 90 degree Celsius, what is the weight of water present?

**Solution:**

To solve this problem, we need to apply the principle of conservation of energy. The energy gained by the steam will be equal to the energy lost by the water. We can calculate the energy gained by the steam using the formula:

Q = m * c * ΔT

Where:
Q is the heat energy gained by the steam,
m is the mass of the steam,
c is the specific heat capacity of steam, and
ΔT is the change in temperature of the steam.

Similarly, we can calculate the energy lost by the water using the formula:

Q = m * c * ΔT

Where:
Q is the heat energy lost by the water,
m is the mass of the water,
c is the specific heat capacity of water, and
ΔT is the change in temperature of the water.

Since the steam is condensing into water, the change in temperature for the steam is from 100 degree Celsius to 90 degree Celsius, which is a decrease of 10 degree Celsius. The change in temperature for the water is from 20 degree Celsius to 90 degree Celsius, which is an increase of 70 degree Celsius.

Now, let's substitute the values into the equations:

Energy gained by the steam:
Q = m * c * ΔT
Q = m * c * -10

Energy lost by the water:
Q = m * c * ΔT
Q = 22 * 4.18 * 70

Since the energy gained by the steam is equal to the energy lost by the water, we can equate the two equations:

m * c * -10 = 22 * 4.18 * 70

Simplifying the equation:

-10m = 22 * 4.18 * 70

m = (22 * 4.18 * 70) / -10

m ≈ -641.32 grams

The negative sign indicates that the mass of the water has decreased. However, it is not physically possible for the mass of water to decrease. Therefore, the steam is not completely condensing into water.

Hence, the weight of water present is still 22 grams.