In a metal chloride, the weight of metal and chlorine is in the ratio of 1:2. The equivalent weight of the metal will be:

(A) 71

(B) 35.5

(C) 106.5

(D) 17.75

To determine the equivalent weight of the metal in the metal chloride where the weight ratio of metal to chlorine is 1:2, we can break down the problem step by step. This involves understanding the concept of equivalent weight and how it relates to the composition of the compound.

Understanding Equivalent Weight

The equivalent weight of a substance is defined as the mass of that substance that combines with or displaces one mole of hydrogen atoms (or 8 grams of oxygen, or 35.5 grams of chlorine, etc.). For metals, the equivalent weight can often be calculated using the formula:

Equivalent Weight = Atomic Weight / n

where n is the number of electrons lost or gained in a reaction.

Analyzing the Metal Chloride

In this case, we have a metal chloride with a weight ratio of metal to chlorine as 1:2. This means that for every 1 part by weight of the metal, there are 2 parts by weight of chlorine. Let's denote the atomic weight of the metal as M and the atomic weight of chlorine as approximately 35.5 g/mol.

Setting Up the Ratio

From the ratio, we can express the weights as:

• Weight of metal = M
• Weight of chlorine = 2 * 35.5 = 71 g

Thus, we can write the ratio as:

M : 71 = 1 : 2

Calculating the Atomic Weight of the Metal

From the ratio, we can deduce that:

M = 71 / 2 = 35.5 g/mol

Finding the Equivalent Weight

Now, to find the equivalent weight of the metal, we need to determine how many electrons it typically loses when forming the chloride. In most metal chlorides, metals lose one electron to form a +1 ion or two electrons to form a +2 ion. Assuming the metal in this case forms a +1 ion (which is common for many metals), we have:

n = 1

Now we can calculate the equivalent weight:

Equivalent Weight = Atomic Weight / n = 35.5 / 1 = 35.5 g

Final Answer

Therefore, the equivalent weight of the metal is 35.5, which corresponds to option (B).