A crucible of 5g weighed 10g after filling with fresh soil. it is then heated in an oven at 100°C for 1 hour. After cooling in a desiccator, the weight was 8g.The percentage of water in the soil is
80%
60%
20%
40%
Explanation
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mass of wet soil =10g-5g=5g
mass of dry mass=8g-5g=3g
mass of water =wet -dry=5g-3g=2g
water in wet soil= 2g/5g times 100=40
My school, the answer is B.
You have to subtract the mass of the crucible from both samples.
Without the mass of the crucible before heating = 5gm
After heating without the mass of the crucible = 3gm
Percentage of water would then be:
3/5 * 100 = 60%.
Please make corrections.
Thanks.
mass of crucible =5gm
freash soil=5gm
mass of crucible+freash soil=10gm
mass of crucible+heated soil(when soil is heated the water evaporate)=8gm
mass of water evaporated=10gm-8gm=2gm
percentage of water evaporated=2/5*100%=40%
D is the correct option
08125861743
Amaziah victor
Taraba state
To determine the percentage of water in the soil, follow these steps:
1. **Calculate the initial weight of the soil sample:**
- Weight of crucible + soil = 10 g
- Weight of crucible alone = 5 g
- Initial weight of soil = 10 g - 5 g = 5 g
2. **Calculate the weight of the soil after heating:**
- Weight of crucible + dried soil = 8 g
- Weight of crucible alone = 5 g
- Weight of dried soil = 8 g - 5 g = 3 g
3. **Calculate the weight of the water lost:**
- Weight of water = Initial weight of soil - Weight of dried soil
- Weight of water = 5 g - 3 g = 2 g
4. **Calculate the percentage of water in the soil:**
- Percentage of water = (Weight of water / Initial weight of soil) × 100
- Percentage of water = (2 g / 5 g) × 100 = 40%
So, the percentage of water in the soil is:
**D. 40%**
percentage of water in the soil =
mass of dry soil - mass of roasted soil / mass of dry soil * 100
